since the above conjecture is wrong in general, I would like to know (and maybe prove) that the following Statement holds:
Let $A,B$ be two closed, convex sets in $\mathbb{R}^n$ such that $A\cap B=\text{bd}(A)\cap \text{bd}(B)\neq\emptyset$ (where $\text{bd}$ denotes the boundary). Then $\text{cone}(A\cap B)=\text{cone}(A)\cap \text{cone}(B)$. (cone is the conic hull operator)
So far I found out that $\text{cone}(A\cap B)\subseteq\text{cone}(A)\cap \text{cone}(B)$ is clear. So lets consider an element $x\in\text{cone}(A)\cap \text{cone}(B)$. So there exisits $a\in A$, $b\in B$, $\lambda,\mu>0$ such that $x=\lambda a$ and $x=\mu b$. Hence $a\in A\cap \text{cone}(B)$ and $b\in B\cap \text{cone}(A)$.
All I need to show the conjecture is to show that $a\in A\cap B$ or $b\in B\cap A$ since this would prove $x\in\text{cone}(A\cap B)$.
If $\text{int}(A)\cap\text{int}(B)\neq\emptyset$ this is not possible I guess, so what about $\text{int}(A)\cap\text{int}(B)=\emptyset$ (int denotes the interior). Do you have an idea?
Edit: The statement should follow immediately if we assume the conic representation of $x$ to be unique. This should be true if $A,B$ are simplices for instance.